ETL-拉鍊算法-帶刪除的拉鍊算法

truncate table CUST;truncate table TAG_CUST;
truncate table vt_inc;truncate table vt_new;
insert into CUST values(100,'張三','13333333333');
insert into CUST values(200,'李四','12222222222');
insert into CUST values(300,'王五','15555555555');
insert into vt_new
    select CUST_ID,CUST_NAME,PHONE_NUM,'20160101','29991231' from CUST;
insert into vt_inc
    select a.* from vt_new a
        left join TAG_CUST b on a.CUST_ID=B.CUST_ID and b.ETL_END_DT='29991231'
        where b.CUST_ID is null or a.PHONE_NUM<>b.PHONE_NUM;
insert into vt_inc
    select b.CUST_ID,a.CUST_NAME,a.PHONE_NUM,a.ETL_START_DT,'19990101' from vt_new a
        right join TAG_CUST b
            on a.CUST_ID=b.CUST_ID
             where a.CUST_ID is null and b.ETL_END_DT='29991231';
update TAG_CUST a set ETL_END_DT='20160102'
    where ETL_END_DT='29991231' and exists
    (select 1 from vt_inc b
     where a.CUST_ID=b.CUST_ID);
insert into TAG_CUST
    select a.* from vt_inc a where a.ETL_END_DT='29991231';